Math 4 part 1

Lesson 4 The Other Circular Functions In Lesson 3, you learned about two circular functions of an angle θ, sine and cosine.Aside from these two functions, there are four other circular functions of an angle θ instandard position. These are the tangent function, cotangent function, secant function andcosecant function.The Tangent Function The third basic function is the tangent function (abbreviated as tan). This function isdefined in terms of sine and cosine functions. Y The tangent function is defined as the set of all X P(x, y) ordered pairs θ, y  where x ≠ 0, θ is an angle in  x 1 θ standard position and y and x are the second and first coordinates of the point of intersection of the terminal side of θ with the unit circle, respectively. Since y = sin θ and x = cos θ, then tan θ = sinθ or y , where cos θ ≠ 0. cos θ xThe Cotangent Function The cotangent function (abbreviated as cot) is defined as the set of all ordered pairs θ, x  where y ≠ 0, θ is an angle in standard position and x and y are the first and second ycoordinates of the point of intersection of the terminal side of θ with the unit circle,respectively. Since x = cos θ and y = sin θ, then cot θ = cos θ or x , where sin θ ≠ 0. sinθ yThe Secant Function The secant function(abbreviated as sec) is defined as the set of all ordered pairsθ, 1  where x ≠ 0, θ is an angle in standard position and x = cos θ. x 19

Since x = cos θ, then sec θ = 1 or 1 , where cos θ ≠ 0. cos θ xThe Cosecant Function The cosecant function (abbreviated as csc)is defined as the set of all ordered pairs θ, 1  where y ≠ 0, θ is an angle in standard position and y = sin θ. y Since y = sin θ, then csc θ = 1 or 1 , where sin θ ≠ 0. sinθ y Did you notice that tangent and cotangent functions are reciprocal functions? Thesame is true for secant and cosecant functions. The value of the tangent, cotangent, secant and cosecant of special angles can beobtained using their x and y coordinates.Examples:1. Evaluate tan 30o, cot 30o, sec 30o and csc 30o. (Note: 30o = π ) 6Solution: P(30o) =  3 , 1  . Hence, cos 30o = 3 and sin 30o = 1 . 2 2 22 Therefore, by definition of tangent, cotangent, secant and cosecant,tan 30o = 1 3= 3 sec 30o = 1 = 1• 2 = 2 • 3 = 2 3 2 = 1• 2 = 1 • 33 3 3 33 3 32 3 3 2 2 csc 30o = 1 = 1• 2 = 2 = 2 1 11cot 30o = 3 3 2 2 = 3•2=2 3 = 1 21 2 22. Determine tan 45o, cot 45o, sec 45o and csc 45o. (Note: 45o = π ) 4Solution: 20

If P(45o) =  2, 2  . Hence, cos 45o = 2 and sin 45o = 2. 2 2 2 2Therefore, by definition of tangent, cotangent, secant and cosecant, 2 sec 45o = 2=2 2= 2tan 45o = 2 = 2 • 2 = 2 2 = 1 1 = 1• 2 = 2 • 22 2 22 2 2 2 22 2 2 csc 45o = 2=2 2= 2 2 1 = 1• 2 = 2 • 22cot 45o = 2 = 2 • 2 = 2 2 = 1 2 22 2 2 2 2 22 2 Notice that the tan 45o and cot 45o are equal and that sec 45o and csc 45o are alsoequal. It is because the right triangle formed by a 45o-45o-90o is an isosceles right triangle.3. Find tan 60o, cot 60o, sec 60o and csc 60o. (Note: 60o = π ) 3If P(60o) =  1 , 3  . Hence, cos 60o = 1 and sin 60o = 3. 2 2 2 2Therefore, by the definition of tangent, cotangent, secant and cosecant, 3 sec 60o = 1 = 1• 2 = 2 = 2 1 11tan 60o = 2 = 3•2=2 3 = 3 2 1 21 2 csc 60o = 1 = 1• 2 = 2 • 3 = 2 3 2 3 3 33 3 2cot 60o = 1 3= 3 2 = 1• 2 = 1 • 33 32 3 3 2 Observe that tan 60o = cot 30o and sec 60o = csc 30o. Why is it so? It is because 30oand 60o are complementary angles. Hence, their functions are also complementary. 21

The tangent, cotangent, secant and cosecant functions of other angles in the form180on ± 30o, 180on ± 45o and 180on ± 60o can be obtained from the functions of 30o, 45o,and 60o, respectively.4. Evaluate tan 0o, cot 0o, sec 0o, and csc 0o.Solution:P(0o) = (1, 0). Hence cos 0o = 1 and sin 0o = 0.Therefore, by definition, tan 0o = 0 = 0 sec 0o = 1 = 1 1 1 cot 0o = 1 = undefined csc 0o = 1 = undefined 0 0Division by zero is not defined. Hence, cot 0o and csc 0o are not defined.5. Evaluate tan 90o. (Note 90o = π ) 2Solution:If P(90o) = (0, 1). Hence, cos 90o = 0 and sin 90o = 1.Therefore, by definition, tan 90o = 1 = undefined sec 90o = 1 = undefined 0 0 cot 90o = 0 = 0 csc 90o = 0 = 0 1 1Division by zero is not defined. Therefore, tan 90o and sec 90o are not defined.The Circular Functions of other Angles The circular functions of angles which are not multiples of the quadrantal angles andspecial angles can be obtained using a scientific calculator or a table of trigonometricfunctions. In this module, the use of a scientific calculator is encouraged for you to make 22

use of the technology. Caution is given that before you use a scientific calculator, that is,you have to familiarize yourself with the model of the scientific calculator you are going touse. This module will not prescribe a particular scientific calculator.Examples:Use a scientific calculator to evaluate the following1. sin 15o 3. cot 100o 5. sec 20o05’2. cos 34o15’ 4. tan 125o40’ 6. csc 320oAnswers:1. sin 15o = 0.258819 4. tan 125o40’ = -1.3933572. cos 34o15’ = 0.826590 5. sec 20o05’= 1.0647433. cot 100o = -0.176327 6. csc 320o = -1.555724 Note that the symbol ‘ means minutes. Thus, 34o15’ means 34 degrees 15 minutes.Answers in the examples are given up to six decimal places.Try this outA. Find the six circular functions of the following angles using the definition of the functions.1. 120o 6. 225o2. 135o 7. 240o3. 150o 8. 270o4. 180o 9. 300o5. 210o 10. 330oB. Find the value of the following using a scientific calculator or a trigonometric table.1. sin 23o 6. csc 102o 11. sec 120o30’ 16. cos 37o25’2. cos 34o 7. tan 44o23’ 12. cot 87o50’ 17. sin 200o52’3. tan 16o 8. cos 48o16’ 13. cos 95o15’ 18. cot 312o45’4. cot 43o 9. sin 55o20’ 14. tan 112o47’ 19. tan 300o35’5. sec 95o 10. cot 29o29’ 15. csc 50o10’ 20. sec 320o28’C. Identify the quadrant/quadrants where the angle is located given the following conditions: 23

1. tan θ > 0 4. cot θ < 0 and cos θ > 02. cot θ < 0 5. sec θ > 0 and sin θ > 03. tan θ > 0 and sin θ < 0 6. cot θ > 0 and sin θ < 0 Let’s Summarize1. If OA (1, 0) is the initial side of an angle on the unit circle and P is any point on the unit circle then, OP is the terminal side of ∠AOP and the coordinates of point P satisfy the equation of the unit circle, x2 + y2 = 1.2. The coordinates of an angle in the form 180°n ± 30° are  ± 3 ,± 1  . 2 23. The coordinates of an angle in the form 180°n ± 60° are  ± 1 ,± 3  . 2 24. The coordinates of an angle in the form 180°n ± 45° are  ± 2 ,± 2  . 2 25. Sine function is the relation between an angle and the y-coordinate while cosine function is the relation between an angle and the x-coordinate.6. Tangent function is the ratio of y to x while cotangent function is the ratio of x to y, where x and y are the coordinates of the point of intersection of the terminal side of an angle in standard position and the unit circle.7. Secant function is the reciprocal of the cosine function while cosecant function is the reciprocal of the sine function. 24

What have you learned1. The y-coordinate of an angle in the along the unit circle is - 4 . If the terminal side of 5 the angle is located in the third quadrant, what is its x-coordinate?a. 3 b. 1 c. - 1 d. - 3 5 5 5 52. An angle measuring 60o is in standard position along the unit circle. What are itscoordinates?a.  1, 3  c.  3 , 1   2 2  2 2b.  − 1 ,− 3  d.  − 3 , 1  2 2 2 23. What are the coordinates of the point of intersection of the terminal side of a 600oangle and the unit circle?a.  3 , 1  c.  2, 2  2 2 2 2b.  − 1 ,− 3  d.  1 ,− 3   2 2  2 24. What is the x-coordinate of a 540o angle along the unit circle?a. − 1 b. - 3 c. - 2 d. -1 2 2 25. It is the reciprocal of the cosine function.a. Tangent function c. Cosecant functionb. Sine Function d. Secant function6. What is sec 240o 25

a. 2 b. 2 3 c. -2 d. − 2 3 3 3 26

Answer KeyHow much do you know1. c 3. b 5. c 7. a 9. a2. c 4. c 6. b 8. d 10. aTry this outLesson 1A. 1. ∠BOA 7.  − 1, 3  2. Right triangle  2 2  3. Pythagorean Theorem  1 ,− 3 4. 3 8.  − 2 2  2 9.  1 ,− 3   2 2 5. positive6.  1 , 3   2 2  10. Pos: QI & QIV; Neg: QII & QIII 11. Pos: QI & QII; Neg: QIII & QIVB. 1. Lies on the unit circle 2. Lies on the unit circle 3. Does not lie on the unit circle 4. Does not lie on the unit circle 5. Lies on the unit circleC 1. y = ± 3 5. y = ± 2 9. x = 2 6 2 2 5 2. x = ± 21 6. x = 3 10. y = 2 6 5 2 7 3. x = ± 3 11 7. x = − 4 3 10 7 4. y = ± 2 2 8. y = − 7 3 3D. 1. 1 6. 5 9. 3 5 2 13 7 2. -0.8 7. − 7 10. − 119 3. − 12 4 12 13 8. − 22 4. 0 5 5. -1 27

Lesson 2 1. 60o  − 1, 3  8. 60o  − 1, 3  15. 30o  − 3 , 1   2 2   2 2   2 2  2. 30o  − 3 , 1  9. 30o  − 3 , 1  16. 45o  − 2, 2   2 2   2 2   2 2  3. 45o  − 2 ,- 2  10. 45o  − 2 ,- 2  17. 60o  1 , 3   2 2   2 2   2 2  4. 60o  − 1,- 3  11. 45o  2, 2  18. 60o  − 1,- 3   2 2   2 2   2 2  5. 60o  1 , - 3  12. 60o  − 1,- 3  19. 60o  − 1, 3   2 2   2 2   2 2  6. 45o  2 ,- 2  13. 45o  − 2 ,- 2  20. 60o  1 , 3   2 2   2 2   2 2  7. 30o  3 , - 1  14. 30o  − 3 , - 1   2 2   2 2 Lesson 3A Terminal Degree Coordinates Sin θ Cos θ Point Measure of Points 1 A (θ) (1, 0) 0 B 2 0o  2, 2  2 2 C 45o  2 2  2 D 0 90o (0, 1) 1 -2 135o  − 2, 2  2 2  2 2  2 E 180o (-1, 0) 0 -1 F 225o  − 2 ,- 2  - 2 -2  2 2  2 2 G 270o (0, -1) -1 0 H 315o  2 ,- 2  -2 2  2 2  2 2 28

B. 1. 3 6. 3 C. 1. I, II 2. II, III 2 2 3. I, IV 4. III, IV 2. − 1 7. − 2 5. II 6. I 2 2 7. IV 8. III 3. − 2 8. 0 2 9. 0 10. 3 4. 1 2 2 5. 0Lesson 4A.Angle Coordinates sin θ cos θ tan θ cot θ sec θ csc θ1. 120o -2  − 1 , 3  3 − 1 −3 − 3 232. 135o 2 2 2 2 3 −2 33. 150o ( )− 2 , 2 2 - 2 -1 -1 -2 3 24. 180o 2 2 2 2 35. 210o 1 2  − 3 , 1  2 -3 − 3 −3 -16. 225o 2 2 2 3 undefined7. 240o 0 (-1, 0) -1 undefined 08. 270o  − 3 ,- 1  -1 - 3 33 - 23 -2 2 2 2 2 3 3 ( )− 2 , - 2 -2 -2 −2 -2 2 2 22 11  − 1 , - 3  - 3 − 1 33 -2 - 23 2 2 2 2 3 3 (0, -1) -1 0 undefined -1 undefined 09. 300o  1 , - 3  -3 1 -3 -3 2 -2 310. 330o 2 2 2 2 -3 3 3 23  3 , - 1  -1 3 3 -3 3 -2 2 2 2 2B. 1. 0.390731 6. 1.022341 11. -1.979294 16. 0.794238 2. 0.829038 7. 0.978703 12. 0.037834 17. -0.356194 3. 0.286745 8. 0.665665 13. -0.091502 18. -0.924390 4. 1.072369 9. 0.822475 14. -2.380844 19. -1.692031 5. -11.47371 10. 1.768694 15. 1.302234 20. 1.296589C. 1. I, III 2. II, IV 3. III 4. IV 5. I 6. IIIWhat have you learned1. d 6. c2. a 7. c3. b 8. a4. d 9. b5. d 29

Module 2 Exponential Functions What this module is about This module is about the roots of exponential equations and zeros ofexponential functions. As you go over this material, you will develop the skills infinding the roots of exponential equations and the zeros of the exponentialfunctions using the property of equality for exponential equation and the laws ofexponents. What you are expected to learn The module is designed for you to use the laws on exponents to find theroots of exponential equations and the zeros of exponential functions How much do you know1. If 3x = 34, what is x?2. Find x if 2x – 1 = 4.3. Simplify the expression (4x5)2.4. Express  4x−4 −2 without negative exponent.  −2x−9 5. Solve for x in the equation 22(5x + 1) = 5006. What are the values of x in 3x2 = 92x−1 ?7. Determine the zeros of the exponential function F(x) = 2x.8. Find the zeros of h(x) = 2x - 3.  1 2 x +1 29. Where will the graph of y = 1 + cross the x-axis?10. What value of x will make the function value of y = 32x – 1 equal to 0?

What you will do Lesson 1 The Property of Equality for Exponential Equation An exponential equation in one variable is an equation where theexponent is the variable. In solving exponential equations, the Property of Equality for ExponentialEquation is used. It is stated as“If a, b, and c are real numbers and a ≠ 0, then, ab = ac if and only if b = c.”Examples: Solve for the value of the variable that would make the equation true.1. 3x = 35 Since the bases are equal, x=5 the exponents must be equal too.Since the x = 5, then 35 = 352. 42y = 48 Since the bases are equal, 2y = 8 the exponents must be equal too. y=4Since y = 4, then 42y = 48 42(4) = 48 48 = 483. 72 = 7z – 1 The bases are equal, 2=z–1 the exponents must be equal too. z=3Since z = 3, then 72 = 7z – 1 72 = 73 – 1 72 = 72Try this outFind the value of the variable that would make each equation true.Set A 1. 3x = 39 2. 23y = 212 3. 52z = 53 4. 8a – 2 = 87 2

5. 12m + 3 = 1206. 44b + 1 = 457. 75x + 2 = 74x – 18. 92(x – 1) = 93x + 19. 64(2x – 1) = 63(x – 1)10. π x2 +2x = π −1Set B 1. 6x = 611 2. 95y = 915 3. 126z = 123 4. 10a – 4 = 10-9 5. 72b + 2 = 7b – 3 6. 23n – 7 = 2n + 5 7. 34(x + 2) = 35x – 2 8. 8– (3 – 2x) = 83(-4 + x) 9. 4x2 − x = 412 10. 54x2 −4x = 5Set C 1. 42x = 49 2. 38y = 312 3. 611z + 2 = 63z – 10 4. 24y + (2 – y) = 23 – 5y5. 8 x = 856. 5 x+3 = 527. 7x2 = 72x−18. 9x2 = 9x+69. π 2x2 −x = π 3( ) ( )2x2+5x 310. 2 =2 3

Lesson 2Review of the Laws of ExponentsLet us review the Laws of Exponents for easy reference.For any real numbers a and b, and any positive real numbers m and n,a. aman = am + nb. (am)n = amnc. (ab)n = anbnd=. am am−n , a ≠ 0 ane. =ab n an ,b ≠ 0 bnf. a0 = 1Examples: Simplify each expression. Express answers with positive exponents: 1. (x3)(x6) (x3)(x6) = x3 + 6 = x9 2. (x -3)2 (x -3)2 = x(-3)(2) = x-6 =1 x6 3. (2x -5)-3 (2x -5)-3 = 2-3x15 = x15 23 = x15 8 4

4. 16x5 12x7 16x5 = 16 ⋅ x5 12x7 12 x7 = 4 ⋅ x5−7 Simplify 3 Express rational exponents = 4 ⋅ x−2 in radical form and simplify. 3 =4⋅ 1 3 x2 =4 3x 2 2  1 3  64x3 5.  −2  125x 3  22  1 3 2  1 3  64x3   64  3  x3    = 125  ·   125 2   2  x − 3 − 3 x 2 2  =  64 ·  1 + 2 3 3 x3 3   125    2 = 42 ·  33 5  x3    [ ]=16  · 2  25 x1 3 2 = 16x 3 25 5

Try this outSimplify each expression. Express answers with positive exponents:Set A 1. (x6)(x7) 2. (3x4)(-6x7) 3. (x5)4 4. (3x3)2 5. 8x9 2x5 6. −15x−2 25x5 7. 21x6 −14x −4 ( )−2x5 2 8. ( )−4x2 39.  8x9 −2  2x5  ( )−2x 2 3 −1 10.  ( ) −2    3x −3Set B 1. (5x-4)(-x2) 2. (-2x-3)(-4x-2) 3. (3x-4)3 4. (-5x-4)2 5. 12x8 20x3 6. −33x−6 24x 2 7. 48x9 −80x −13 6

( )−2x−4 28. ( )3x2 −39.  − 6x5 −2  8x9  ( ) 4x2 −1  −2 10.  −3  ( )   −2x −2Set C 1. −27x6 21x 42. (2x2 )3 8x93. (3x−2 )4 (9x3 )−24. (8x6 )2 (2x3 )4 2  1 35.  8x 2   27x   16.  64x−6 2  36x2 7. −  32x9 −2  −72x −13 ( )8. 2x3 −3 2   ( ) 2   −3x 29.  − 2x −2  −2 8x4    7

(( ))10.−2x 2−2  −3 −3   − 2x −2  Lesson 3 Finding the Roots of Exponential Equation The Property of Equality for Exponential Equation which is also known asEquating-Exponents Property implies that in an exponential equation, if thebases are equal, the exponents must also be equal.Examples: Solve each exponential equation. 1. 24x + 2 = 8x – 2 Use laws of exponents to make the bases equal. Then apply the EquatingExponents Property. 24x + 2 = 8x – 2 24x + 2 = 23(x – 2) 4x + 2 = 3(x – 2) 4x + 2 = 3x – 6 x = -8 2. 9-x = 1 27 Use laws of exponents to make the bases equal. Then apply the EquatingExponents Property. 9-x = 1 27 32(-x) = 27-1 3-2x = (33)-1 3-2x = 3-3 8

-2x = -3 x= −3 −2 x= 3 23. 2x2 −5x = 1 16 Use laws of exponents to make the bases equal. Then apply the EquatingExponents Property. 2x2−5x = 1 16 2x2 −5x = 16−1 ( )2x2 −5x = 24 −1 2x2 −5x = 2−4 x2 – 5x = -4 x2 – 5x + 4 = 0 (x – 1)(x – 4) = 0 x = 1 or x = 4Try this outSolve for x in each exponential equation.Set A 1. 2x = 128 2. 5x = 125 3. 2x = 1 2 4. 92x = 27 5. 2x – 1 = 32 6. 1000x = 1002x – 5 7. 43x + 1 = 8x – 1 8. 93x = 27x – 2 9

9. 8x – 1 = 163x 10. 93x + 1 = 273x + 1Set B 1. 9x = 27 2. 4x = 128 3. 4x = 1 2 4. 3x + 2 = 27 5. 25x + 1 = 125x 6. 4x = 0.0625 7. 3– x = 1 243 8. 8x + 3 = 1 16  1 − x 3 9. = 27 2 x +1 10.  1 3−2x = 8−x 4Set C 1.  1 x = 4 2 2. 4x = 1 8 3. 5x + 3 = 25 4. 9x = 27x + 1 5. 73x + 2 = 49x 6. 1213x = 11x – 1 x+3 7. 3 3 = 815+x  16 x2 + x 25 8. =1 ( )9.5x −2 = 0.04 3 10

1 10.  1 3 x = 1 8 2 Lesson 4 Determining the Zeros of Exponential Functions As in other function, the zero of an exponential function refers to the valueof the independent variable x that makes the function 0. Graphically, it is theabscissa of the point of intersection of the graph of the exponential function andthe x-axis. To find the zero of an exponential function f(x), equate f(x) to 0 and solvefor x.Examples: Determine the zero of the given exponential function. 1. f(x) = 2x To find the zero of the function, equate it to 0 and solve for x. f(x) = 2x = 0 2x = 0 The resulting equation suggests that f(x) has no zero since no real value of x will make 2x = 0 a true statement. 2. g(x) = 34x – 8 – 1 To find the zero of the function, equate it to 0 and solve for x. g(x) = 34x – 8 – 1 = 0 34x – 8 – 1 = 0 34x – 8 = 1 34x – 8 = 30 4x – 8 = 0 4x = 8 x=2 11

The zero of g(x) is 2.  1 2 x +3 33. h(x) = +9 To find the zero of the function, equate it to 0 and solve for x.  1 2 x+3 3 h(x) = −9 =0  1 2 x+3 3 −9 =0  1 2x +3 3 = 9  1 2x +3 3 = 32 ( )3−1 2x+3 = 32 ( )3 −2x−3 = 32 -2x – 3 = 2 -2x = 5 x= −5 2 The zero of h(x) is − 5 . 2  1 x −1 5124. y = 22x + 3 – To find the zero of the function, equate it to 0 and solve for x. y = 22x + 3 –  1 x−1 = 0 512 22x + 3 –  1 x−1 = 0 512 12

 1  x −1 512 22x + 3 = ( )22x + 3 = 512−1 x−1 ( )22x + 3 = 29 −1 x−1   ( )22x + 3 = 2−9 x−1 ( )22x + 3 = 2 −9x+9 2x + 3 = -9x + 9 11x = 6 x= 6 11 The zero of y is 6 . 11Try this outDetermine the zeros of the given polynomial function.Set A 1. f(x) = 3x 2. g(x) = 3x + 2 - 27 3. h(x) = 4x – 1 – 64 4. p(x) = (7-x)2 – 343 5. F(x) = (0.2)x – 2 – 54x 6. G(x)= 82x – 22x + 1  1 x −3 27. H(x) = − 81−x8. P(x) = 4x(x – 5) –  1 3 16( )9. y = 27 x−3 − 33x−1 13

 2 4 x +3 − 16 3 8110. y =Set B 1. f(x) = 22x – 1 2. g(x) = 165 + x – 4 3. h(x) = 103x – (0.001)x + 34. p(x) = 2x(x – 3) – 1 165. F(x) =  1 3 x − 4 − 23−2x 646. G(x) = 343 – 492x – 1( )7. H(x) = 3 x −1  1  x −2 5 125 −  2 x −3 32 3 2438. P(x) = −  3 2 x +1 − 27 5 1259. y =  4 3x −1  49  x 7 1610. y = −Set C 1. f(x) = 92x + 2 – 27  1 x −3 52. g(x) = 1253x + 7 –3. h(x) = 25x + 1 – 82x – 34. p(x) = 275x – 6 – 97x + 35. x 1 F(x) = 814 − 276. G(x) = (9x)(3x) – 243  3 x +5 729 2 647. H(x) = − 14

 4 2 x −3 27 3 648. P(x) = −9. y =  2 3 x +1 − 625 5 1610. y =  3 x−3 − 4 2 9 Let’s summarize1. An exponential equation is an equation where the variable is the exponent.2. In solving exponential equations, the Property of Equality for Exponential Equation is used. Stated as follows: “If a, b, and c are real numbers and a ≠ 0, then, ab = ac if and only if b = c.”3. For any real numbers a and b, and any positive real numbers m and n, a. aman = am + n b. (am)n = amn c. (ab)n = anbn d=. am am−n ,a ≠ 0 an e. =ba n an ,b ≠ 0 bn f. a0 = 14. The Property of Equality for Exponential Equation which is also known as Equating-Exponents Property implies that in an exponential equation, if the bases are equal, the exponents must also be equal.5. The zero of an exponential function refers to the value of the independent variable x that makes the function 0.6. Graphically, the zero of an exponential function is the abscissa of the point of intersection of the graph of the exponential function and the x-axis. 15

7. To find the zero of an exponential function f(x), equate f(x) to 0 and solve for x. What have you learned1. If 22x + 1 = 25, what is x?2. Find x if 4x + 1 = 2.3. Simplify the expression (5x4)3.4. Express  9x−3 −2 without negative exponent.  −36x7 5. Solve for x in the equation 9(42x + 1) = 36x6. What are the values of x in 4x2 = 82x−1 ?7. Determine the zeros of the exponential function F(x) = 5x – 2.8. Find all the zeros of f(x) = 7x – 3 – 49.9. Where will the graph of g(x) =  1 x –  1  2 x +1 cross the x-axis? 4 210. What value of x will make h(x) = 322x –  1 −2x equal to 0? 16 16

Answer Key Set B 1. 11How much do you know 2. 3 3. 1 1. 4 2 2. 3 4. -5 3. 16x10 5. -5 4. 4 6. 6 7. 10 x10 8. 9 5. 2 9. 4 and -3 6. 2 ± 2 10. 2 ± 2 7. no zero 2 8. 5 9. 0 and 3 Set C 1. 9 2 2 10. 0 2. 3 2Try this out 3. - 3 2Lesson 1 4. 1 8Set A 5. 25 1. 9 6. 1 2. 4 7. 1 3. 3 8. 3 and -2 5 9. 3 and -1 4. 9 2 5. -3 10. 1 and -3 6. 1 2 7. -3 8. -3 17 9. 1 5 10. -1

Lesson 2 7. − 3x22 5Set A 1. x13 8. 108 2. -18x11 x2 3. x20 4. 9x6 9. 16x8 5. 4x4 9 6. − 3 5x7 10. x16 7. − 3x10 4 2 8. − x4 Set C 4 1. − 9x2 9. 1 7 16x8 10. − 1 2. 1 72 x3Set B 3. 6561 1. − 5 x2 x2 2. 8 4. 4 x 3. 27 5. 4 x 1 x12 3 4. 25 x8 9 5. 3x5 5 6. 4 6. − 11 3x4 8x8 7. - 81 16 x 31 8. 1 5184 x 26 9. 1 4 x12 10. - x20 8 18

Lesson 3 9. - 3 5Set A 1. 7 10. 6 2. 3 7 3. -1 4. 3 Set C 4 1. -2 5. 6 2. - 3 6. 10 2 7. - 1 3. -1 5 4. -3 8. -2 5. -2 9. - 1 6. - 1 3 5 10. - 1 7. - 57 3 11 8. 0 and -1Set B 9. 3 1. 3 10. -1 2 2. 7 Lesson 4 2 3. - 1 Set A 2 1. No rational zero 4. 1 2. 1 5. 2 3. 4 6. -2 4. - 3 7. 5 2 8. - 13 5. 2 3 5 6. 1 4 7. 0 8. 2 and 3 19

9. - 7 7. 1 3 8. 0 9. - 5 10. 1 4 3 10. 1Set B How much have you learned 1. No rational zero 1. 2 2. - 9 2. - 1 2 3. - 3 2 2 3. 125x12 4. 3 ± i 7 4. 16x20 2 5. -1 5. - 27 6. 3 ± 3 16 6. 7 2 8 7. none 7. 7 8. 5 11 9. It will not cross the x-axis 8. 8 10. 0 9. 1 10. 1 20 5Set C 1. - 1 4 2. - 9 5 3. 10 4. 24 5. -3 6. 5 3

Module 2 Linear Functions What this module is about This module is about linear function of the form f(x) = mx + b. As you goover this material, you will develop the skill in determining different aspects oflinear function such as slope, trend, intercepts and some points that belong to thegraph of the linear function. It is also expected that you will develop the skill informing linear functions of the form f(x) = mx + b, given certain conditions. What you are expected to learn This module is designed for you to: 1. determine the following: a. slope b. trend (increasing or decreasing) c. x- and y-intercepts; and d. some points given f(x) = mx + b 2. determine f(x) = mx + b given: a. slope and y-intercept b. x- and y-intercept c. slope and a point; d. any two points How much do you know 1. What is the slope of the f(x) = 2x – 3? 2. True or false? The trend of a linear function with negative slope is increasing. 3. Solve for the y-intercept of 3x – 2y = 6. 4. Which of the following step will solve for the x-intercept of the linear function y = 4x + 16?

a. Substitute zero to x and solve for y. b. Substitute zero to y and solve for x. c. Take the square root of the constant term. d. Take the negative reciprocal of the coefficient of x. 5. What is the x-intercept of y = 4x + 16? 6. Name three points on the graph of y – 3x + 2 = 0. 7. Determine the linear function with slope 5 and y-intercept -2. 8. The x- and y-intercepts of a linear function are both 3. Find the linear function. 9. The graph of a linear function passes through the point (-4, 0), and its slope is -2. Express the linear function in the form f(x) = mx + b. 10. What linear function has (2, 4) and (-3, 1) as points on its graph? What you will do Lesson 1 Determining the Slope, Given f(x) = mx + b In the linear function f(x) = mx + b, m is the slope and b is the y-intercept.Examples: Determine the slope and the y-intercept of each linear function. 1. f(x) = 3x + 8 The slope is 3 and the y-intercept is 8. 2. y = 2x – 5 The slope is 2 and the y-intercept is -5. 3. g(x) = -4x + 7 The slope is -4 and the y-intercept is 7. 4. y = 1 x 2 The slope is 1 and the y-intercept is 0. 2 5. 3x + 2y = 12 First, transform the equation 3x + 2y = 12 in of form f(x) = mx + b 2

3x + 2y = 12 2y = -3x + 12 -3x is added to both sides =2y −3x + 12 Divide both sides by the coefficient of y 2 22 y =− 3 x + 6 The fractions are simplified. 2 The slope is − 3 and the y-intercept is 6. 2Try this outDetermine the slope and the y-intercept of each linear function.Set A1. f(x) = 4x + 52. g(x) = -7x + 13. h(x) = 4x4. F(x) = 3 – 4x5. 2x + 7y = 14Set B1. F(x) = 8x – 92. G(x) = -4x – 63. H(x) = x4. 3x + 8y = 245. 51x – 17y – 4 = 1Set C1. f (x)= x + 1 22. g(=x) 9 x − 3 423. h(x) = -0.3x + 0.14. f(x) = -5.2x – 4.45. -9x + 8y – 12 = 0 Lesson 2 Determining the Trend, Given f(x) = mx + b The trend of a linear function is said to be increasing if the slope ispositive. 3

The trend of a linear function is said to be decreasing if the slope isnegative.Examples:Determine the trend of each linear function.1. f(x) = 8x + 3 The slope is 8; hence, the trend of the function is increasing.2. y = 5x – 2 The slope is 5; hence, the trend of the function is increasing.3. g(x) = -7x + 4 The slope is -7; hence, the trend of the function is decreasing.4. y = 1 x 2The slope is 1 ; hence, the trend of the function is decreasing. 25. 4x + 3y = 12 First, transform the equation 4x + 3y = 12 in of form f(x) = mx + b. 4x + 3y = 12 3y = -4x + 12 -4x is added to both sides =3y −4x + 12 Divide both sides by the coefficient of y 3 33 y =− 4 x + 4 The fractions are simplified 3The slope is − 4 , hence, the trend of the function is decreasing. 3Try this outDetermine the trend of each linear function.Set A1. f(x) = 5x + 42. g(x) = -x + 73. h(x) = -4x – 54. F(x) = 4 – 3x5. -5x + 3y + 15 = 0Set B1. F(x) = 9x – 82. G(x) = -5x + 1 4

3. H(x) = -5 – x4. 8x + 3y = 245. 17x – 51y – 4 = -1Set C1. h(x) = -0.1x + 0.32. 5x – 6 = y3. 3x + y = 114. -8x + 12y – 9 = 05. 25x – 15y – 3 = 2 Lesson 3 Determining the X-and Y-intercepts, Given f(x) = mx + b To determine the y-intercept of a linear function, substitute x = 0 and solvefor the value of y. This value is the y-intercept. To determine the x-intercept of a linear function, substitute y = 0 and solvefor the value of y. This value is the x-intercept.Examples:Determine the x- and the y-intercepts of each linear function.1. f(x) = 3x + 8 0 is substituted to y Let y = f(x) = 3x + 8. -8 is added to both sides 0 = 3x + 8 Apply symmetric property of equality -8 = 3x Both sides are divided by 3 3x = -8 0 is substituted to x x = −8 The numerical expression is simplified 3 The x-intercept is − 8 . 3 y = 3(0) + 8 y=8The y-intercept is 8.2. 3x + 2y = 12 0 is substituted to y 3x + 2(0) = 12 Divide both sides by the coefficient of x 3x = 12 3x = 12 33 x=4 5

The x-intercept is 4. 0 is substituted to x 3(0) + 2y = 12 Divide both sides by the coefficient of y 2y = 12 2y = 12 22 y=6The y-intercept is 6.Try this outDetermine the x- and the y-intercepts of each linear function.Set A1. f(x) = 8x + 162. g(x) = 4x – 23. y = 3 – 4x4. 2x – 5 = y5. -7x + 7y = 14Set B1. F(x) = -14x + 72. 2y = -4x – 63. 4x + 6 = 2y4. 16x – 12 = 4y5. -12x + 22y – 14 = 0Set C1. 6x – 30 = 18y2. -32x + 8y – 12 = 03. 28x – 12y – 3 = 14. G(x) = -0.3x + 0.15. y = -5.2x – 4.4 Lesson 4Determining Some Points of a Given Linear Function To determine a point on the graph of a given linear function y = f(x):A. Get any element from the domain of the linear function. Recall that the domain of any linear function is the set of real numbers. This element is the abscissa of the point.B. Substitute this abscissa to x in the linear function. 6

C. Solve for the corresponding value of y. This value is the ordinate of the point.D. Repeat steps A to C using other element from the domain of the linear function to determine other points on the graph of the linear function.Examples:Determine three points on the graph of each linear function.1. y = x + 4 If x = 0, then Step A y=0+4 Step B y=4 Step CThe point (0, 4) is on the graph of y = x + 4. If x = 2, then y=2+4 y=6The point (2, 6) is also on the graph of y = x + 4. If x = -5, then y = -5 + 4 y = -1The point (-5, -1) is also on the graph of y = x + 4.2. 3x – 4y = 36 If x = 0, then Step A Step B 3(0) – 4y = 36 Step C 0 – 4y = 36 -4y = 36 −4y = 36 −4 −4 y = -9The point (0, -9) is on the graph of 3x – 4y = 36. If x = 2, then 3(2) – 4y = 36 6 – 4y = 36 -4y = 30 7

−4y = 30 −4 −4 y = − 15 2The point  2, − 15  is also on the graph of 3x – 4y = 36. 2 If x = -4, then 3(4) – 4y = 36 12 – 4y = 36 -4y = 24 −4y = 24 −4 −4 y = -6The point (-4, -6) is also on the graph of 3x – 4y = 36.Try this outDetermine three points on the graph of each linear function.Set A1. f(x) = x + 62. g(x) = -2x + 73. h(x) = 2x4. y = 4 – 3x5. -2x + y = 4Set B1. g(x) = 5x – 42. 2y = -2x – 63. 4x + 6 = 2y4. -12x + 22y – 14 = 05. 14x – 21y – 4 = 3Set C1. y = -7x2. y= 2− x 43. 28x – 12y – 3 = 14. f (=x) 9 x − 3 425. h(x) = -0.3x + 0.1 8

Lesson 5Determining f(x) = mx + b, Given the Slope and Y-intercept To determine f(x) = mx + b, given the slope and y-intercept of the linearfunction, substitute the given slope to m and the given y-intercept to b.Examples: Determine f(x) = mx + b, given the slope and y-intercept of the linearfunction.1. slope = 2, y-intercept 3Since the slope is 3, m = 2 and the y-intercept is 3, b = 3. Hence,f(x) = mx + bf(x) = 2x + 32. slope = -3, y-intercept 1Since the slope is -3, m = -3 and the y-intercept is 1, b = 1. Hence,f(x) = mx + bf(x) = -3x + 13. slope = 1, y-intercept − 3 2Since the slope is 1, m = 1 and the y-intercept is − 3 , b = − 3 . Hence, 22f(x) = mx + bf (x=) (1)x +  − 3  2f (x)= x − 3 24. slope = 4 , y-intercept 0 3Since the slope is 4 , m = 4 and the y-intercept is 0, b = 0. Hence, 33f(x) = mx + bf (=x) 4 x + 0 3f (x) = 4 x 3 9

Try this out Determine f(x) = mx + b, given the slope and y-intercept of the linear function. Set A 1. slope 3, y-intercept 2 2. slope -2, y-intercept 4 3. slope 5, y-intercept -7 4. slope -8, y-intercept -1 5. slope 6, y-intercept 0 Set B 1. slope 1, y-intercept -1 2. slope -9, y-intercept 0 3. slope 4 , y-intercept -2 3 4. slope − 1 , y-intercept − 1 33 5. slope 7, y-intercept − 4 3 Set C 1. slope 2 , y-intercept − 1 34 2. slope − 2 , y-intercept − 1 53 3. slope 0.2, y-intercept 1 4. slope 1, y-intercept -0.1 5. slope 2, y-intercept 2 Lesson 6 Determining f(x) = mx + b, Given the X - and Y- intercepts Let y = f(x). To determine f(x) = mx + b, given the x- and y-intercept of thelinear function, substitute the given x-intercept to a and the given y-intercept to bin the form of the equation x + y =1 aband then, solve for y in terms of x. 10

Examples:Determine f(x) = mx + b, given the x- and y-intercept of the linear function.1. x-intercept 2, y-intercept 3Since the x-intercept is 2, a = 2 and the y-intercept is 3, b = 3. Hence, x + y =1 ab x + y =1 23 6  x + y  =6 (1) Multiply both sides by the LCD, 6  2 3  -3x is added to both sides 3x + 2y = 6 Both sides are divided by 2 2y = -3x + 6 =2y −3x + 6 2 22 y =− 3 + 3 2 f (x) =− 3 + 3 22. x-intercept = -3, y-intercept = 1, hence, x + y =1 a b x + y =1 −3 1 −3 x + y  =−3(1) Multiply both sides by the LCD, –3 −3 1  x – 3y = -3 -3y = -x – 3 x is subtracted from both sides =y −x − 3 Both sides are divided by –3 −3 −3 =y 1 x + 1 3 f (=x) 1 x + 1 33. x-intercept 1, y-intercept − 3 2Since the x-intercept is 1, a = 1 and the y-intercept is − 3 , b = − 3 . Hence, 22 11

x + y =1 ab x + y =1 1 −3 2    − 3  x + y  =− 32 (1) Multiply both sides by the LCD, − 3 2  1 −3  2 2 3 x is added to both sides 2 − 3 x + y =− 3 22=y 3 x − 3 22 f (=x) 3 x − 3 224. x-intercept 4 , y-intercept − 6 35Since the x-intercept is 4 , a = 4 and the y-intercept is − 6 , b = − 6 . 33 55Hence, x + y =1 ab x + y =1 4 −6 35 3x − 5y =1 46 12  3x − 5y  =12 (1) Multiply both sides by the LCD 12.  4 6  9x – 10y = 12 -10y = -9x + 12 9x is subtracted from both sides=y −9x + 12 Both sides are divided by –10 −10 −10=y 9 x − 6 10 5 f=(x) 9 x − 6 10 5Try this out Determine f(x) = mx + b, given the slope and y-intercept of the linearfunction. 12

Set A 1. x-intercept 3, y-intercept 2 2. x-intercept -2, y-intercept 4 3. x-intercept -1, y-intercept -3 4. x-intercept -8, y-intercept -1 5. x-intercept -4, y-intercept -8 Set B 1. x-intercept 5, y-intercept 6 2. x-intercept -4, y-intercept 12 3. x-intercept -12, y-intercept -16 4. x-intercept -15, y-intercept -18 5. x-intercept − 1 , y-intercept 1 37 Set C 1. x-intercept 13, y-intercept 1 5 2. x-intercept − 1 , y-intercept 3 9 3. x-intercept − 8 , y-intercept -3 3 4. x-intercept − 4 , y-intercept 1 72 5. x-intercept 2 , y-intercept 13 9 12 Lesson 7 Determining f(x) = mx + b, Given the Slope and One Point Let y = f(x). To determine f(x) = mx + b, given the slope and one point ofthe linear function, substitute the given slope to m and the coordinates of thegiven point to x1 and y1 in the form of the equation y – y1 = m(x – x1) and then,solve for y in terms of x.Examples: Determine f(x) = mx + b, given the slope and one point of the linear function. 1. slope = 2, (1, 3) Since the slope is 2, m = 2 and the given point is (1, 3), x1 = 1 and y1 = 3. 13

Hence,y – y1 = m(x – x1) Distributive property is appliedy – 3 = 2(x – 1) 3 is added to both sidesy – 3 = 2x – 2y = 2x – 2 + 3y = 2x + 12. slope = -3, (2, -1) Since the slope is -3, m = -3 and the given point is (2, -1), x1 = 2 and y1 = -1. Hence,y – y1 = m(x – x1) Distributive property is appliedy – (-1) = -3(x – 2) 1 is subtracted from both sidesy + 1 = -3x + 6y = -3x + 6 – 1y = -3x + 5 3. slope 1 , (-1, -4) 2 Since the slope is 1 , m = 1 and the given point is (-1, -4), x1 = -1 and y1 = 22-4. Hence,y – y1 = m(x – x1) y − (−4) = 1 [x − (−1)] 2 y + 4 = 1 (x +1) 22(y + 4) = 2 1 (x + 1) Both sides are multiplied by 2 2 Distributive property is applied 8 is subtracted from both sides2y + 8 = x + 1 Both sides are divided by 22y = x + 1 – 82y = x – 7 2y = x − 7 22 y= 1x−7 224. slope − 4 , (−2, 3) 3 14

Since the slope is − 4 , m = − 4 and the given point is (-2, 3), x1 = -2 and 33y1 = 3. Hence, y – y1 = m(x – x1) (Distributive property is applied) (3 is added to both sides) y − 3 = − 4 [x − (−2)] (Renaming 3) 3 y − 3 = − 4 (x + 2) 3 y−3 = − 4 x − 8 33 y = −4x − 8 +3 33 y=−4x−8+9 3 33 y=−4x+1 33Try this outDetermine f(x) = mx + b, given the slope and a point of the linear function.Set A1. slope 3, (2, 5)2. slope 4, (3, -4)3. slope 2, (-5, 1)4. slope -2, (-4, -2)5. slope 5, (-7, 0)Set B1. slope 5, (-2, 2)2. slope -3, (3, 1)3. slope 1, (6, -1)4. slope -9, (-5, 0)5. slope 1 (2, −3) , 3Set C1. slope 2 , (4, −1) 32. slope − 1 ,  4, 1  2  2 15

3. slope − 2 ,  2 , 1  3 3 2( )4. slope 0.2, 1, 0.3( )5. slope −0.3, 0.1, 0.2 Lesson 8 Determining f(x) = mx + b, Given Any Two Points Let y = f(x). To determine f(x) = mx + b, given any two points of the linearfunction, substitute the coordinates of the first point to x1 and y1, and thecoordinates of the second point to x2 and y2 in the form of the equation y − y1 = y2 − y1 (x − x1 ) x2 − x1and then, solve for y in terms of x.Examples:Determine f(x) = mx + b, given two points of the linear function.1. (2, 1), (3, 4) Since the first point is (2, 1), x1 = 2 and y1 = 1 and the second point is (3,4), x2 = 3 and y2 = 4. Hence, y − y1 = y2 − y1 (x − x1 ) x2 − x1 y −1 = 4 −1 (x − 2) 3−2 y −1 = 3 (x − 2) 1 y – 1 = 3(x – 2) y – 1 = 3x – 6 y = 3x – 6 + 1 y = 2x – 52. (-2, 5), (0, -3) Since the first point is (-2, 5), x1 = -2 and y1 = 5 and the second point is (0,-3), x2 = 0 and y2 = -3. Hence, 16

y − y1 = y2 − y1 (x − x1 ) x2 − x1 y − 5 = − 3 − 5 [x − (− 2)] 0 − (−2) y − 5 = − 8 (x + 2) 2 y – 5 = -4(x + 2) y – 5 = -4x – 8 y = -4x – 8 + 5 y = -4x – 33. (-2, -7), (-3, -1) Since the first point is (-2, -7), x1 = -2 and y1 = -7 and the second point is (-3, -1), x2 = -3 and y2 = -1. Hence, y − y1 = y2 − y1 (x − x1 ) x2 − x1 y − (−7) = −1 − (−7) [x − (− 2)] − 3 − (−2) y − 7 = 6 (x + 2) −1 y – 7 = -6(x + 2) y – 7 = -6x – 12 y = -6x – 12 + 7 y = -6x – 5Try this outDetermine f(x) = mx + b, given two points of the linear function.Set A1. (1, 3), (-2, 6)2. (3, 0), (0, -2)3. (-5, 0), (-4, 7)4. (0, 4), (4, 0)5. (0, 1), (1, -2)Set B1. (1, 4), (5, 6)2. (8, -2), (6, -4)3. (9, 0), (5, -2)4. (5, -6), (6, -5)5. (9, 3), (5, -6) 17

Set C 1. (-3, -2), (2, 1) 2. (-1, 3), (1, 0) 3. (-2, -3), (3, -1) 4. (-1, 3), (3, -3) 5. (-1, -1), (-3, -3) Let’s summarize1. In the linear function f(x) = mx + b, m is the slope and b is the y-intercept.2. The trend of a linear function is said to be increasing if the slope is positive. The trend of a linear function is said to be decreasing if the slope is negative.3. To determine the y-intercept of a linear function f(x) = y, substitute x = 0 and solve for the value of y. This value is the y-intercept. To determine the x-intercept of the linear function, substitute y = 0 and solve for the value of y. This value is the x-intercept.4. To determine a point on the graph of a given linear function y = f(x): a. Get any element from the domain of the linear function. Recall that the domain of any linear function is the set of real numbers. This element is the abscissa of the point. b. Substitute this abscissa to x in the linear function. c. Solve for the corresponding value of y. This value is the ordinate of the point. d. Repeat steps a to c using other element from the domain of the linear function to determine other points on the graph of the linear function.5. To determine f(x) = mx + b, given the slope and y-intercept of the linear function, substitute the given slope to m and the given y-intercept to b.6. To determine f(x) = y = mx + b, given the x- and y-intercept of the linear function, substitute the given x-intercept to a and the given y-intercept to b in the form of the equation x + y =1 ab and then, solve for y in terms of x.7. To determine f(x) = y = mx + b, given the slope and one point of the linear function, substitute the given slope to m and the coordinates of the given 18

point to x1 and y1 in the form of the equation y – y1 = m(x – x1)and then,solve for y in terms of x.8. To determine f(x) = y = mx + b, given any two points of the linear function, substitute the coordinates of the first point to x1 and y1, and the coordinates of the second point to x2 and y2 in the form of the equationy − y1 = y2 − y1 (x − x1 ) x2 − x1and then, solve for y in terms of x. What have you learned1. What is the slope of the f(x) = -5x + 4?2. True or false? The trend of a linear function with positive slope is increasing.3. Solve for the y-intercept of 6x – 3y = 4.4. Which of the following step will solve for the y-intercept of the linear function y = 3x + 9? a. Substitute zero to x and solve for y. b. Substitute zero to y and solve for x. c. Take the square root of the constant term. d. Take the negative reciprocal of the coefficient of x.5. What is the x-intercept of 12 = 9y + 18x?6. Name three points on the graph of y – 2x + 4 = 0.7. Determine the linear function with slope -3 and y-intercept -4.8. The x- and y-intercepts of a linear function are both -2. Find the linear function.9. The graph of a linear function passes through the point (-3, 7), and its slope is -4. Express the linear function in the form f(x) = mx + b.10. What linear function has (-2, -5) and (-3, 1) as points on its graph? 19